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I'm trying to make an audio amplifier for an 8 ohm speaker. I'm new in electronics area and having a problem with it. I already had other issues before arriving where I am, like the need of a high power stage because the low impedance of the load.

To the high power stage my tentative after some reading was to use a transformer at the transistor collector to match the low impedance of the speaker with the high impedance of the amplifier output.

The first problem were, if I use an 9V DC source for biasing the transistor, the voltage at transformer primary will float at maximum in 0 to 9V (In an extreme case). So, if the speaker can work at a maximum peak to peak voltage of for example 6V, the ratio n1/n2 should be at the maximum 1.5, because the peak voltage at secondary will be 9/1.5 = 6V. Is that a right reasoning?

Going ahead, if I use a transformer with 1.5:1, the impedance seen in the primary will be 8*(1.5^2) = 18 ohm. So the configuration of this stage of the amplifier would be like this:

Schematic showing a BJT and a transformer on the collector

The transistor is biased with 4.5V at emitter resistor, 4.5V at collector-emitter (Vce), and the voltage at the primary is almost 0V for DC. The capacitor at emitter has low impedance and it's almost a short circuit for AC. So, in my mind, the voltages will vary from 0V in the primary and 9V at Vce, to 9V in primary and 0V at Vce (in an extreme case).

But when I run it, these waveforms appears. Yellow curve represents the input voltage (250 mV peak), and purple is the voltage at the primary of the transformer (not the output). It have a positive peak of 130mV, but at the bottom seems not to be what I expected. The curve is being cut out at -20mV. It doesn't should saturate only in a relative high voltage, next to the 9V of CC source?

What is wrong and how can I solve this?

EDIT:

I have changed the bias of the transistor for a collector current of 50mA, with a base resistor of 27k instead of 1.35M. Also removed the emitter capacitor (which really reduced the distortion) and changed the emitter resistor to 10 ohm. The clipping has went away for these low input voltage (250 Vpk), and the voltage at primary is apparently a perfect senoid, with amplitude a little higher than the input voltage. But if I increase the input voltage to 3V peak, the clipping reappears at the bottom. I still can't figure out why this happens.

EDIT 2:

I'm adding the image of the waves for the last update:

the clipping voltage has increased, but it is still happening. It starts to appear at about 1V peak at the input.

And I don't know if I have made it clear but this is the end stage of my amplifier, with a low gain, only for drive the speaker. I'm making the stages separately for easy understanding of each part. As this stage has a high input impedance I will use a common emitter before for high gain of voltage.

And sorry for my bad english :(

EDIT 3 (solved):

The problem was the transistor must be biased with a collector DC current of at least the same value that the peak to peak current which load requires. Other thing I noticed is that the BC548 I was using fails with a relative big current (>200mA) even in the simulation, the base-emitter voltage increases significantly, also making the wave to clip.

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  • \$\begingroup\$ You are not saying a lot about your design goals (distortion, freq response, power efficiency etc). It might be worthwhile looking at a class AB amp with capacitive coupling to the speaker. Lots of good internet info on this. \$\endgroup\$ – RoyC Dec 21 '16 at 22:22
  • \$\begingroup\$ You don't need the transformer but you do need something significantly more robust than a BC548. \$\endgroup\$ – user207421 Dec 21 '16 at 22:25
  • \$\begingroup\$ My purpose is to make this for learning and for while my goal is mainly to make it works without clip the signal and not drop the input voltage. I've looked at class AB for power stage and it should be a best alternative, but I want to understand why what I'm trying isn't working instead of leave it. \$\endgroup\$ – glz Dec 22 '16 at 1:14
  • \$\begingroup\$ Due to Core saturation it must be large to apply 9V DC across primary and leakage will be high. MUST Use Centre tap transformer to reset flux. roetta.it/ik3hia/transistor/Transistors_diagrams/… \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 22 '16 at 20:22
  • \$\begingroup\$ @glz Do you understand my 2nd answer yet? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 26 '16 at 2:28
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Second answer: getting back to basics

  1. You cannot analyze anything completely, unless all impedances are known.
    • Learn what impedance is Z(f,dc), how to measure and compute it for every component and looking in & out of any branch
  2. This is just a simulation, BUT what is real and what is ideal and what is time scale steady state or ? e.g. is LdI/dt steady state?
    • e.g what is transformer Inductance, hFE of Q1, What is Z(f)?
  3. What is the basic configuration "common emitter" What does ideal common emitter do?

    • amplify base current with infinite collector impedance at source
    • voltage gain is Impedance Ratio *I = Collector/Emitter * Current.
    • What is load Z(f) on collector ?
      • Np/Ns = n, turns ratio, n² * R_Load or Ic= Iout/n
  4. What is gain from base to emitter current?

    • Is AC current less than DC current so Ie does not drop to zero?

      1. Does Vce drop to zero?
    • then AC current exceeds DC current bias.

    • then output appears to "clip". (your problem)
    • except wrong name. Clipping is when Ic goes to 0 with max Vce.
    • Here Vce goes to 0 thus we call it "saturated"
  5. Why saturated?

    • because 8 Ohm AC Load draws more current than Collector DC current
    • Ip=Vp/Rload Idc > Ip is mandatory to keep Ic>0.
    • thus for 8Vp out Idc=8V/8Ω = 1A then Ic must be >2Adc to supply AC 2App This is crucial to understand for Class A
    • Remember this is a Class A amplifier
    • What is overall gain of Class A
    • Vin to Ie AC gain Ie= Ve/Re , for AC Vb=Ve (if Vbe>0.6)
    • Vc = Ic * Rc for high hFE Ic=Ie thus
    • Av = (Rc/Re)
    • collector load = Rc = ωL // 8Ω , if ωL>> 8Ω Rc=8Ω
    • Thus overall gain = Rc/Re = 8/10 assuming 1:1 XFMR
  6. Why is a single Rb to V+ poor?

    • DC collector current depends on hFE which can be >+/-50%
  7. How would Rbase be improved if it connects to collector for negative feedback?

    • it lowers output impedance and input impedance
    • it regulates gain better instead of being a current source
    • now it becomes a voltage gain device with R ratio Av= Rf/Rin
      • where Rf is the collector to base feedback
      • and Rin is source series impedance.

Here is what it takes to drive 8 Ω with 8Vp or 16Vpp in Class A, using negative feedback with Av<10. Note the difference from your values as the impedance ratio is essential to proper Dc bias, gain and avoidance of saturation, clipping and significant distortion. Due to efficiency Class AB is far better and Class E even better.

Note the Java Sim link above shows scope +/- peak readings (snap below) enter image description here

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  • \$\begingroup\$ Thank you very much! But I have some questions. 1)The problem seems to be that I was biasing the transistor with a DC current smaller than the speaker needs in AC peak to peak, right? 2)You have not used the tap of the transformer, the problem of the core saturation will stay as we have a big DC current on the primary? \$\endgroup\$ – glz Dec 26 '16 at 16:39
  • \$\begingroup\$ 1) yes 2) transformer saturation is not a problem on simulator. ( my bad. ) only in real life. Since Class A uses twice the current (pk-pk) vs push-pull on centre tap where each driver only handle 1/2 wave but biased to overlap smooth. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 27 '16 at 1:07
  • \$\begingroup\$ Taking advantage of your answer, what is the function of the capacitor in parallel with the load? \$\endgroup\$ – glz Dec 27 '16 at 12:44
  • \$\begingroup\$ can you guess, knowing Zc(f)=1/2piCf optional \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 27 '16 at 19:32
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Two major things wrong. First the resistor values in your circuit are way too high to properly have a decent standing DC current from collector through emitter. Yes you need a DC current to permanently bias the transistor into the correct part on the load line and this will need tens if not hundreds of mA.

The 2nd problem is one of understanding - any inductor be it a normal inductor or the primary of a transformer cannot have a sustained DC voltage across it and, as such, it will cause the collector signal voltage (when you get the biasing right as per point 1) to rise and fall equally above and below 9 volts. Potentially, with a perfect transistor (rarer than unaffordium and costlier than unobtainium) you can get 18 volts p-p from a 9 volt DC power source.

Thirdly, as a hint, get rid of the emitter capacitor - you'll never ever be satisfied with distortion figures for this type of class A amplifier with too much capacitance there even when you get the biasing right (hint look at the emitter resistor being more like ten ohms).

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  • \$\begingroup\$ 'Cannot have DC across it'? \$\endgroup\$ – user207421 Dec 21 '16 at 22:25
  • \$\begingroup\$ @EJP ok to be precise, cannot sustain a dc voltage across it. Does that fit better for you and, because it's xmas, I shall update my answer to correct my sloppiness. \$\endgroup\$ – Andy aka Dec 21 '16 at 22:57
  • \$\begingroup\$ Ok thanks for the response! I have changed the bias of the transistor for a collector current of 50mA, with a base resistor of 27k instead of 1.35M. Also removed the emitter capacitor (which really reduced the distortion) and changed the emitter resistor to 10 ohm. The clipping has went away for these low input voltage (250 Vpk), and the voltage at primary is apparently a perfect senoid, with amplitude a little higher than the input voltage. But if I increase the input voltage to 3V peak, the clipping reappears at the bottom. I still can't figure out why this happens. \$\endgroup\$ – glz Dec 22 '16 at 1:03
  • \$\begingroup\$ How can you get 18v p-p using one transistor? Wouldn't you need two and drive both ends of the transformer differentially? \$\endgroup\$ – gbarry Dec 22 '16 at 1:23
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    \$\begingroup\$ @gbarry try simulating it and see. The quiescent pont at the collector has to be 9V else there will be a monstrous DC current flowing through the primary and transistor limited only by the primary winding resistance. The voltage rises above and below 9 volts equally. \$\endgroup\$ – Andy aka Dec 22 '16 at 10:10
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@glz simple Common emitters Class A circuits dont work well driving a transformer to 8 ohm speaker due to the DC primary current core saturation. Also single stage transistor amps are good for voltage gain or current gain but not both with high power gain. which you are trying to do.

So this will never work extremely well. Something will always be compromised. THD, efficiency, input impedance, etc etc. saturation of core, saturation of Vce. excessive loading on 9V battery, poor efficiency

Use a complementary emitter follower output stage with pre-amp stages for voltage gain of 20~ 50.

you MUST use a centre tap audio transformer in order to reverse the flux in the core and not saturate it with DC even though it is biased with DC. The DC current must be enough to overcome the load AC current to prevent starving the complmentary driver .

enter image description here This is taken from a 60's vintage pocket AM radio

Turns ratio may be something like 10:1 not 1.5:1

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  • \$\begingroup\$ I have not let clear but this is only the last stage of the amplifier, I will use a common emitter before for provide a decent gain of voltage. \$\endgroup\$ – glz Dec 22 '16 at 12:55
  • \$\begingroup\$ I've pressed return to drop a line but it posted the comment. I will see about emitter follower and read more about core saturation to see if it clarifies anything. \$\endgroup\$ – glz Dec 22 '16 at 13:00
  • \$\begingroup\$ @glz study the above perfect 7 transistor AM radio design. You may reverse polarity to NPN. They may have used Germanium PNP \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 22 '16 at 20:58
  • \$\begingroup\$ comments questions? the lack of centre tap is your answer. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 23 '16 at 17:18
  • \$\begingroup\$ Thanks for the tip, but where do I connect the center tap to reverse the flux? Is it at the primary or the secondary of the transformer? \$\endgroup\$ – glz Dec 24 '16 at 0:28

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